已知数列{an}中,an=(2n-1)2^n,求前n项和sn
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已知数列{an}中,an=(2n-1)2^n,求前n项和sn
已知数列{an}中,an=(2n-1)2^n,求前n项和sn
已知数列{an}中,an=(2n-1)2^n,求前n项和sn
运用错位相减法来解答!
an=a1+a2+a3a4+……+an
an=1*2^1+3*2^2+5*2^3+7*2^4+……+(2n-1)2^n①
2an= 1*2^2+3*2^3+5*2^4+……+(2n-3)*2^n+(2n-1)2^(n+1)②
①-②,-an=2+2*2^2+2*2^3+2*2^4+……+2*2^n-(2n-1)2^(n+1)=2+2(2^2+2^3+2^4+……+2^n)-(2n-1)2^(n+1)=(3-2n)2^(n+1)-6